If y = x^2, then find the zeros of the polynomial P(x) = x^2+y^2+2xy+1

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embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

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Find the zeros of `P(x)=x^2+y^2+2xy+1` if `y=x^2` :

Substitute for `y` to get:





Setting `P(x)=0` we solve for `x` :



`x(x+1)=+-i`  (Take square root of both sides; `i=sqrt(-1)` )

`x^2+x+-i=0`  Using the quadratic formula we get:



P(x) has no real zeros. The complex zeros are:







sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

The problem provides the information that `y = x^2` , hence, you need to substitute `x^2`  for y in polynomial P(x) such that:

`P(x) = x^2 + (x^2)^2 + 2x*x^2 + 1`

`P(x) = x^4 + 2x^3 + x^2 + 1`

You need to find the zeroes of polynomial, hence, you should solve the equation `P(x) = 0`  such that:

`x^4 + 2x^3 + x^2 + 1 = 0`

Notice that the expression `x^4 + 2x^3 + x^2 + 1`  will be larger than zero for any value of x.

Hence, evaluating the zeroes of the given polynomial yields that P(x) has no real zeroes.


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